Question

Suppose we wish to estimate the area\\ \(\int_0^\pi \frac{\cos(x)}{x^2} dx\)\ using Simpson's rule with 2n subintervals. Furthermore, suppose we want to make sure that the error in our estimate is such that\\ \(e_s(2n) \le 1 \times 10^{-3}\)\ What is the smallest value of 2n required to do this?\\ To work this our you should first calculate\\ \(K_4 = \max_{x \in I} |f^{(4)}(x)|.\)\ What is\\ K = \\ From this you should now be able to calculate this smallest value of 2n. What is it?\\ 2n =

          Suppose we wish to estimate the area\\
\(\int_0^\pi \frac{\cos(x)}{x^2} dx\)\
using Simpson's rule with 2n subintervals. Furthermore, suppose we want to make sure that the error in our estimate is such that\\
\(e_s(2n) \le 1 \times 10^{-3}\)\
What is the smallest value of 2n required to do this?\\
To work this our you should first calculate\\
\(K_4 = \max_{x \in I} |f^{(4)}(x)|.\)\
What is\\
K = \\
From this you should now be able to calculate this smallest value of 2n. What is it?\\
2n =
        
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Suppose we wish to estimate the area

∫0^π(cos(x))/(x^2) dxusing Simpson's rule with 2n subintervals. Furthermore, suppose we want to make sure that the error in our estimate is such that

es(2n) ≤ 1 × 10^-3What is the smallest value of 2n required to do this?

To work this our you should first calculate

K4 = maxx ∈ I |f^(4)(x)|.What is

K = 

From this you should now be able to calculate this smallest value of 2n. What is it?

2n =

Added by Noelia H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Suppose we wish to estimate the square root of 2 to 10 decimal places. What is the smallest value of n required to do this? To work this out, you should first calculate K = m * n * f. What is K? From this, you should now be able to calculate the smallest value of 2n. What is it? 2n =
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Transcript

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00:01 So based in the given statement, you can express that as 2n, it's equivalent to 1 half.
00:10 So something for the value of n, we'll just have two.
00:14 Divide both sides by 2...
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